Question

In the figure, the bisectors of $\angle A$ and $\angle B$ meet at a point. If $\angle C={100}^{\circ }$ and $\angle D={50}^{\circ },$ Find the measure of $\angle APB$.

Answer 1

In quadrilateral ABCD,

$\angle D={50}^{\circ },\angle C={100}^{\circ }$
PA and PB are the bisectors of $\angle A$ and $\angle B$.
In quadrilateral ABCD,
$\angle A+\angle B+\angle C+\angle D={360}^{\circ }$
(Sum of angles of a quadrilateral)
$\Rightarrow \angle A+\angle B+{100}^{\circ }+{50}^{\circ }={360}^{\circ }\angle A+\angle B+{150}^{\circ }={360}^{\circ }\angle A+\angle B={360}^{\circ }={150}^{\circ }={210}^{\circ }$
and $\frac{1}{2}\angle A+\frac{1}{2}\angle B=\frac{{210}^{\circ }}{2}={105}^{\circ }$
$(\because$ PA and PB are bisector of $\angle A$ and $\angle B$ respectively )
$\angle PAB+\angle PBA={105}^{\circ }\Rightarrow \angle PAB+\angle PBA+\angle APB={180}^{\circ }$
(Sum of angles of a triangle)
$\Rightarrow {105}^{\circ }+\angle APB={180}^{\circ }\Rightarrow \angle APB={180}^{\circ }-{105}^{\circ }={75}^{\circ }\therefore \angle APB={75}^{\circ }$